tegwin
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| posted on 10/3/09 at 09:17 AM |
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Acceleration and Maths
Lying in bed last night I had a kinda dumb moment....
What do you guys recon...
If a car accelerates from 0-60 in 6 seconds
How long would it take to get to 30?
Simplicity says that its half the speed, so half the time... But I dont think this is the case due to the fact the car is accelerating.... or is
it?
(for arguments sake lets just say we have an "average car" 
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jlparsons
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| posted on 10/3/09 at 09:19 AM |
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If the car is accelerating uniformly, it'll be half the time I think. But normally they don't cos of gears and powerbands and stuff.
Caveat: I think.
[Edited on 10/3/09 by jlparsons]
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omega0684
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| posted on 10/3/09 at 09:22 AM |
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it would only take half the time if acceration was constant.
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matt_claydon
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| posted on 10/3/09 at 09:25 AM |
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To 30 you'll be in first gear, 30-60 will be in second (unless it's a BEC). Therefore accel to 30 will be much quicker.
Also, air resistance increases with the square of speed, so the faster you go the more torque is used to overcome drag and the less is available for
acceleration. Therefore even if you did 0-60 in one gear the acceleration rate would be a decreasing curve.
This assumes a flat torque curve.
[Edited on 10/3/09 by matt_claydon]
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vinny1275
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| posted on 10/3/09 at 09:27 AM |
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If you're going for fastest possible run to 60, not only will you be accelerating, but the rate of acceleration will be increasing as well,
particularly as you have the inertia of the car to get rolling as well, so I'd say 0-30 will be slower than 30-60.
If you're really interested (and can't sleep again tonight), someone like Evo might have their test graphs from their datalogging online,
so you could compare different cars for their run to 30 / 60....
Cheers
Vince
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SteveWalker
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| posted on 10/3/09 at 09:28 AM |
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As others have said, it would be half the time if acceleration were constant, which with a car, it is not.
For general calculations involving time, motion, acceleration, distance and constant acceleration, google suvat.
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02GF74
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| posted on 10/3/09 at 09:29 AM |
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shcool boy maths: v = u + at.
since we are accelerating from standstill, i.e. 0 mph, u = 0.
If acceleration, a, is constant, then it will be 1/2 the time but a won't be constant mainly due to air resistance.
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coozer
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| posted on 10/3/09 at 09:31 AM |
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Don't forget at the start of the run your stationary and at the end 60... there is inertia to overcome so I reckon the last half of the run will
go quicker, say 20mph @ 3 secs, 40 @ 4.5 secs vrooooom...
1972 V8 Jago
1980 Z750
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jlparsons
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| posted on 10/3/09 at 09:36 AM |
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As energy in a moving object is calculated as 1/2 x mass x velocity squared, and as the engine gives a set power output, doesn't that mean
acceleration would decrease according to the inverse square of the speed...?
So... (my head hurts)... slightly less than 3 seconds to 30 assuming power at the rear wheels is constant?
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approval, terms and conditions apply. Apply only to affected area. For recreational use only. All models over 18 years of age. No user-serviceable
parts inside. Subject to change. As seen on TV. One size fits all. May contain nuts. Slippery when wet. For office use only. Edited for television.
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v8kid
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| posted on 10/3/09 at 09:49 AM |
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I'll go with that!
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hellbent345
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| posted on 10/3/09 at 09:50 AM |
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if acceleration was 10 metres per second per second, and we are thinking about reaching say 60m per second (for ease of explaination), it would take
one second to get to 10m per second, 2 seconds to get to 20m per second 3 seconds to get to 30m per second 4 to 40m/s 5 to 50m/s and 6 to 60m/s. This
assumes perfectly linear acceleration though.
maths
a= (v-u)/t
where v is final velocity and u is initial velocity. T is time.
so a= 60-0/6 = 10m/s/s
transposing T=(v-u)/a
T=(30-0)/10 = 3 seconds
But this is purely theoretical, as previously mentioned rolling resistance of tyres and all components in drivetrain, inertiaof the car and all moving
components in drivetrain, air resistance all make this pretty much rubbish lol
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hellbent345
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| posted on 10/3/09 at 09:53 AM |
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quote:
As energy in a moving object is calculated as 1/2 x mass x velocity squared
i think this is kinetic energy rather than energy supplied by an engine or similar. Ie this is the energy the car itself will gain by accelerating to
the final velocity.
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smart51
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| posted on 10/3/09 at 09:55 AM |
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quote:
Originally posted by matt_claydon
air resistance increases with the square of speed, so the faster you go the more torque is used to overcome drag and the less is available for
acceleration. Therefore even if you did 0-60 in one gear the acceleration rate would be a decreasing curve.
The POWER required to overcome air resistance increases with the CUBE of speed. Rolling resistance of your tyres roughly increases with speed.
If your engine has a flat torque curve and you get to 60 in 1st, 0-30 will be done with most of the engine's power. From 30-60, more of the
engine's power is used to overcome drag, so less is available for acceleration.
Normal cars get to 30 or so in 1st and maybe 60 ish in 2nd. Lower gears multiply the engines torque by a greater amount so you have more acceleraton
in lower gears.
The third factor is the engine's torque curve. Here's an oddball example: If your engine revs to 12,000 RPM and will do 60 in 1st but has
no torque at all below 6000 (lets say it is a tiny engine with a huge turbo), 0-30 acceleration will be poor but when the engine comes alive, 30-60
could be great.
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Paul TigerB6
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| posted on 10/3/09 at 09:55 AM |
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If you fancy some fast BEC action, then all the figures are available on the video of the Tiger Z100WR doing the 0-60mph run (in 2.97s!!). It also
shows the accelerative g-force which is quite interesting also. Shame they didnt publish the figures up to 100mph (i believe its 6.6s). Peak
acceleration seems to be pretty much where you would expect it - 1st gear pretty much around where the maximum torque is produced.
linky
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hughpinder
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| posted on 10/3/09 at 10:44 AM |
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As far as I can remember, 0-100 times are usually about twice the 0 to 60 time.
Assuming you have no downforce from your body shape:
The problems are:
from 0 to low speed, geting the power on to the road.
Then you get to a point where you can achieve linear acceleration according to the grip available to the tyres.
Then one of 2 things starts to reduce the acceleration you can achieve: engine power- the gain in kinetic energy to double your speed from some
initial value is 4 times that required to get to the first speed, to treble your speed is 9 times the enery etc...OR you are limited by wind
resistance/aerodynamics which increases as the cube of speed (approximately), or a combination of these two.
For a seven type car, you are unlikely to reach engine power or aerodynamic factors by 60mph, so the only things stopping 30 taking half the time of
60 are inital grip at take off and gear changes.
Hugh
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02GF74
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| posted on 10/3/09 at 11:53 AM |
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quote:
Originally posted by coozer
Don't forget at the start of the run your stationary and at the end 60... there is inertia to overcome so I reckon the last half of the run will
go quicker, say 20mph @ 3 secs, 40 @ 4.5 secs vrooooom...
I don't think so. It has been a while since I did this sort of stuff but.....
F = ma
inertia is related to mass.
the force is what is pushing the car so using the above, a will be the same regardless of the speed of the car. (ignoring any other forces such as air
resistance, friction etc)
All things being equal, let's say engine can supply force F. A car weighing twice as much will end up with 1/2 a so mass (and thus inertia) has
been taken care of.
Look at it another way - inertia is an object's resistance to change in motion i.e. change in speed or direction; the inertia is the same
regardless if the object is staitonary or travelling at 30 mph.
(I think this is correct, been a while.....)
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Rosco
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| posted on 10/3/09 at 12:32 PM |
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Obviously a keen interest in maths so lets add a bit more.
I think the biggest factors would be the engine torque curve, gearing and wind resistance. Don't believe this then just compare how long to
takes to accelerate from 0-30mph vs. 100-130mph.
To calculate it accurately I think you'd need to look work out all the forces at a particular speed - i.e. torque at the wheels (allowing for
gearing), wind resistance, rolling resistance... then divide that by mass to give acceleration against speed. Then you'd simply need to
integrate this against time to give velocity against time. piece of cake?
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02GF74
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| posted on 10/3/09 at 12:45 PM |
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Peeps - we are trying to think too hard about this. 
OP was this.
quote:
Originally posted by tegwin
If a car accelerates from 0-60 in 6 seconds
How long would it take to get to 30?
Simplicity says that its half the speed, so half the time...
correct, v = u + at
But I don't think this is the case due to the fact the car is accelerating.... or is it?
no, "the fact that the car is accelerating" means nothing;
Now if acceleration decreases, then car will take longer to get from 30 to 60 than it would from 0 - 30.
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jlparsons
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| posted on 10/3/09 at 02:25 PM |
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quote:
Originally posted by hellbent345
quote:
As energy in a moving object is calculated as 1/2 x mass x velocity squared
i think this is kinetic energy rather than energy supplied by an engine or similar. Ie this is the energy the car itself will gain by accelerating to
the final velocity.
This is kinetic energy, but that's exactly the same thing as the energy supplied by the engine. For the car to reach a given velocity it must
receive that level of kinetic energy and that energy can only come from the engine. So given that the engine has a maximum power output and that each
additional unit of speed requires the addition of more kinetic energy than the last, acceleration must decrease with speed.
But when we're looking at road-going speed i don't think this is going to have any measurable effect. Perhaps it might come into it if
we're talking bugattis!
[Edited on 10/3/09 by jlparsons]
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hellbent345
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| posted on 10/3/09 at 03:42 PM |
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ah yes i really should have considered where the energy was coming from lol, i think i must have been thinking of energy being conjured from thin air
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